Quick questions on DC and AC motors: HSC Physics Module 6

Consider a rectangular coil of $n$ turns, side lengths $a$ and $b$ (so area $A = ab$), carrying current $I$ in a uniform field $\vec{B}$. The two sides of length $a$ that lie perpendicular to $\vec{B}$ experience forces $F = nBIa$ in opposite directions, forming a couple. The lever arm is $(b/2) \cos \theta_{\text{plane-to-field}}$ on each side, giving a total torque:

If you simply attach a DC supply to a coil in a magnetic field, the torque drives the coil toward the face-on position, decelerating as it approaches and then reversing direction past it. The coil would oscillate about the equilibrium, not rotate.

When the coil rotates in the field, the changing flux through it induces an EMF (Faraday's law). By Lenz's law this induced EMF opposes the supply voltage that is causing the rotation: it is a back EMF $\varepsilon_{\text{back}}$.

AC motors split into two broad families. Both rely on the same idea: produce a rotating magnetic field in the stator (the stationary part) by feeding multi-phase AC into a set of coils arranged around the rotor.

A car starter motor has an armature with $50$-turn coils of area $0.030$ m$^2$. The field strength is $0.50$ T and the supply voltage is $12$ V. The armature resistance is $0.040$ ohms.

The rotor is a magnet (a permanent magnet or an electromagnet fed by slip rings). It locks onto the rotating stator field and spins at exactly the same frequency (the synchronous speed, $f_{\text{rotor}} = f_{\text{supply}}$). Used in clocks, turntables and precision applications.

The rotor is a set of conducting bars short-circuited at each end (no slip rings or commutator at all). The rotating stator field sweeps past the rotor, inducing currents in the bars (Faraday's law). These currents, sitting in the rotating field, experience a magnetic force that drags the rotor in the direction of rotation (Lenz's law: the induced current opposes the change, that is, the relative motion of field past rotor).

The HSC formula sheet typically gives $\tau = nBIA \cos \theta$ with $\theta$ as the angle between the plane of the coil and the field. If you measure $\theta$ from the area normal instead, the formula becomes $\tau = nBIA \sin \theta$. Either is fine if you are consistent; check carefully which one the question expects.

It does not reduce the useful mechanical power. The mechanical power delivered to the shaft equals $\varepsilon_{\text{back}} I$. Back EMF is the channel through which electrical energy becomes mechanical energy.

A split-ring commutator (DC motor) reverses the current direction in the coil every half-turn. Slip rings (AC generator, synchronous motor) deliver an unbroken AC signal to or from the coil.

Without slip there is no relative motion between field and rotor, no induced EMF, and no torque. An induction motor cannot run exactly at synchronous speed.

A DC motor uses a commutator and runs on DC. A standard AC motor runs on AC and either uses slip rings (synchronous) or no electrical contact to the rotor at all (induction).

Like Faraday's law, the turns multiply the effect of a single loop.